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Grex > Science > #63: Power and drag measurements - 1993 _Grand Am_ SE automatic | |
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drew
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Power and drag measurements - 1993 _Grand Am_ SE automatic
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Jul 2 19:55 UTC 2000 |
As Friday was virtually wind-free, or at least as calm as it gets,
it seemed a good time to do the power requirement measurements on my
latest car. The speed range is not as wide as I would have liked; but
the data should be satisfactory for most purposes.
The data was fitted to a third degree polynomial with the data
for 80, 70, 60, and 50 MPH as boundary conditions. The resulting
function was within my reaction time for the other data points; so
I judged the results good to two significant figures.
The calculations file, Lotus 123 format, is in ~drew/grand_am.wk1.
-------
Mass (curb): carpoint.msn.com/Vip/Specifications/Pontiac/Grand%20Am/1993
2756 lb
Crew &
cargo: 150 lb approx
Fuel: 45 lb approx
Total: 2951 lb
Speed Time Rough Poly Accel Accel Drag Power
(MPH) (sec) diff calc (MPH/sec) (grav) (lbf) (W)
------------------------------------------------------------------------
80 7.6 80 -0.81462 -0.03713 -109.585 -17500
75 13.6 6 75.26804 -0.76272 -0.03476 -102.603 -15400
70 20.8 7.2 70 -0.70066 -0.03194 -94.2551 -13200
65 28 7.2 65.17786 -0.63885 -0.02912 -85.9406 -11200
60 36.6 8.6 60 -0.56535 -0.02577 -76.0533 -9100
55 47.1 10.5 54.53279 -0.47610 -0.02170 -64.0464 -7000
50 57.6 10.5 50 -0.38737 -0.01765 -52.1108 -5200
Speed vector: 80 70 60 50
Time matrix: 1 1 1 1
7.6 20.8 36.6 57.6
57.76 432.64 1339.56 3317.76
438.976 8998.912 49027.89 191102.9
Inverse: 2.290999 -0.21251 0.006008 -0.00005
-2.08755 0.367957 -0.01326 0.000130
0.946291 -0.18643 0.008937 -0.00010
-0.14973 0.030991 -0.00168 0.000025
Polynom constants:86.44175 -0.88062 0.004350 -8.0E-07
Deriv constants: -0.88062 0.008701 -2.4E-06
-------------
As expected, the power requirement, at least for one-minute miles,
is much less than that of an _Impala_'s 25 kW.
As for overall efficiency, I've measured 30 miles per gallon under
conditions of a mixture of 70 MPH average freeway and variable surface
speeds. Taking 2.3 gallons per hour as the fuel consumption under
standard freeway operations, and 115000 BTU/gallon of fuel, we get
264500 BTU/hour, which is about 77.5 kW. The efficiency, fuel to
propulsion, works out at around 17%. I seem to recall higher efficiencies
with earlier vehicles.
I'll post results for the station wagon when I manage to get data
for it.
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| 28 responses total. |
rcurl
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response 1 of 28:
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Jul 2 21:07 UTC 2000 |
Probably doesn't matter, but doing a least-squares fit of your polynomial
to all the data points, rather than an exact fit to just four, would
distribute the error and provide a more "efficient" parameter estimation.
Also, if you do an orthogonal polynomial fitting, you can test what
order of polynomial is sufficient.
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drew
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response 2 of 28:
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Jul 3 23:22 UTC 2000 |
Thanks. I'll try it soon as I can find some information on least squares and
orthogonal fitting.
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drew
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response 3 of 28:
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Jul 4 03:59 UTC 2000 |
Matlab does least-squares, an area wherein Lotus 123 seems to be deficient.
I shall have to see about writing an add-on...
Coefficients are slightly different, but not grossly so. I expect the final
results to be very much the same.
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rcurl
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response 4 of 28:
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Jul 4 06:00 UTC 2000 |
In really interesting cases, fitting a third order polynomial to four
points, can give really wild fits. The most important point is to use *all
the data*. An estimator that uses all the data is called an efficient
estimator, and always has the smallest variances of the parameter
estimators. You could illustrate this point by comparing the sum of
squared deviations for the remaining three points when an exact fit is
made to four points, with the sum of squared deviations for a least
squares estimator using all seven points.
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drew
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response 5 of 28:
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Jul 9 02:37 UTC 2000 |
I've got data for the other car now; and I've redone the calculations for the
first one. ~drew/cardrag.wk1 does not include code for doing the least-squares
fit, which I did in MatLab using the POLYFIT function, since I don't know how
to get Lotus 123 to do that yet.
Grand Am:
Mass (curb): carpoint.msn.com/Vip/Specifications/Pontiac/Grand%20Am/1993
2756 lb
Crew &
cargo: 150 lb approx
Fuel: 45 lb approx
Total: 2951 lb
Speed Time Poly Accel Accel Drag Power kW*hr
(MPH) (sec) calc (MPH/sec) (grav) (lbf) (W) /mile
------------------------------------------------------------------------
80 7.6 80.00630 -0.86589 -0.03947 116.5 18624 0.233
75 13.6 75.07199 -0.78044 -0.03557 105.0 15736 0.210
70 20.8 69.78537 -0.69032 -0.03146 92.9 12991 0.186
65 28 65.09877 -0.61376 -0.02797 82.6 10726 0.165
60 36.6 60.15105 -0.54008 -0.02462 72.7 8712 0.145
55 47.1 54.83992 -0.47636 -0.02171 64.1 7044 0.128
50 57.6 50.04656 -0.44146 -0.02012 59.4 5934 0.119
Polynom constants:87.04021 -0.98765 0.008507 -4.4E-05
Deriv constants: -0.98765 0.017014 -1.3E-04
========================================================================
Safari (wagon):
Mass (curb): carpoint.msn.com/Vip/Overview/Pontiac/Safari/1988.asp
4182 lb
Crew &
cargo: 150 lb approx
Fuel: 30 lb approx
Total: 4362 lb
Speed Time Poly Accel Accel Drag Power kW*hr
(MPH) (sec) calc (MPH/sec) (grav) (lbf) (W) /mile
------------------------------------------------------------------------
85 3.7 85.25442 -1.42615 -0.06501 283.6 48174 0.567
80 8.1 79.40030 -1.23836 -0.05645 246.2 39370 0.492
74 12.5 74.32544 -1.07195 -0.04886 213.2 31523 0.426
70 16.8 70.02861 -0.92997 -0.04239 184.9 25870 0.370
65 22.7 65.03736 -0.76838 -0.03502 152.8 19848 0.305
60 30 59.99955 -0.62164 -0.02833 123.6 14822 0.247
55 39 54.92147 -0.52172 -0.02378 103.7 11403 0.207
50 48.6 50.03282 -0.51370 -0.02341 102.1 10207 0.204
Polynom constants:90.84928 -1.60061 0.024596 -0.00018
Deriv constants: -1.60061 0.049193 -5.5E-04
As expected from rough fuel economy measurements, it takes about twice as much
energy, mile for mile and MPH for MPH, to run the station wagon as it does
the smaller car. A station wagon type hull must somehow be better for drag
than a sedan, since somehow it can do one-minute miles on less than 15 bkW
of power, as compared to 25 kW for the car I first did this experiment with
('81 _Impala_). I'm getting about the same fuel economy, but I've also
typically run the station wagon around 70 MPH instead of 60; and the station
wagon *is* a bit more massive - irrelevant on the highway, but a disadvantage
on surface streets.
Re the odd data point in the station wagon test: I was a bit late in calling
off the 75 MPH mark.
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rcurl
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response 6 of 28:
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Jul 9 19:51 UTC 2000 |
Data (and analyses) like these should be inlcuded in car evaluations!
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gull
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response 7 of 28:
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Jul 9 20:07 UTC 2000 |
Hmm. Wonder what the figures would look like for my Econoline. It has all
the aerodynamic design of a boxcar. I don't have the tools to do the curve
fitting, though.
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n8nxf
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response 8 of 28:
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Jul 10 12:30 UTC 2000 |
It took this mathematical lackey a while to figure out what the heck you were
doing but I think I got it. I don't know how valuable this sort of data would
be as there are a lot of variables. Tire type and pressure would be a
significant one that doesn't seem to be addressed here. Same goes for how
well the wheels are aligned. As for it being useful for purchasing a car,
the numbers can be easily muddied by options such as roof-racks, side mirrors,
wheel and body trim options, additional suspension components increasing drag
under the car, the mechanical drag added by going from 2WD to 4WD, etc.
Your drag figures for the station wagon seem a little high or the figures for
the sedan are a little low. I don't see how the drag of the wagon can be over
twice that of the sedan! Two sedans are more efficient than one station
wagon? It just doesn't make sense!
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drew
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response 9 of 28:
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Jul 10 16:27 UTC 2000 |
It's not a full sized sedan. The '93 _Grand Am_ is a smaller car than in
previous years. In fact, a sedan that I tested over a decade ago was *less*
efficient than the current wagon - 25 kW for one-minute miles, compared with
less than 15 kW for the wagon.
As for the variables, yes, that's a problem. Still, I consider this
information better than none at all.
Re #7: I intend to see about teaching Lotus 123 to do curve fitting (have to
look it up myself first). You might want to collect your data during the next
convenient calm period. You'll need a stretch of highway that is both straight
and level; and a tape recorder, along with a sound card and Windows Sound
Recorder to play it into, will be extremely helpful.
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rcurl
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response 10 of 28:
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Jul 10 17:13 UTC 2000 |
There will always be a lot of variables in any comnparison of two cars,
even of identical models. Yet, cars are always being compared! In fact,
the comparisons that are made are seldom as precise as this, in part
because they average driving conditions over some distance. That is
both the strength and weakness of what drew is doing. His measurements
give instantaneous variables, and would have to be also done up and
down hill (for example), etc. Of course, they also don't measure engine
performance - just the power losses. But those are what the engine makes up
for, so this helps separate the contributions of the active (engine)
and passive (vehicle) components.
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russ
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response 11 of 28:
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Jul 12 03:32 UTC 2000 |
Re #7: It's not that hard to do a least-squares fit with a calculator.
I did my first one with a BASIC program I wrote given an algorithm
on the blackboard, IIRC. If I was interested, I could probably
derive the algorithm now (I didn't know calculus then).
Re #9: Instead of a tape recorder, a camcorder might be better. Just
aim it at the speedo, mention significant events for the soundtrack and
do the timing by frame-counting on your VCR at home. (This won't work
if your speedometer is a bit sticky, like mine.)
If you can also see the outside in the FOV, the camcorder also allows
you to associate acceleration anomalies with the terrain.
Re #10: Once power demand is determined, it's not much of a jump to
convert fuel consumption into an overall drivetrain efficiency. This
can be very illuminating.
One of the complications not mentioned thus far is that crosswinds
will increase vehicle drag in both directions along a road. This
might, or might not, make a significant change in the numbers derived.
The obvious way to control for it is to test on a calm day.
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rcurl
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response 12 of 28:
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Jul 12 05:40 UTC 2000 |
What's still missing is a fuel consumption rate meter.
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drew
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response 13 of 28:
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Jul 12 16:48 UTC 2000 |
That would be nice to have. I considered getting one, but they were scarce;
and the one I did see in an auto parts store (calibrated in "miles per gallon
at 35 MPH and 55 MPH), as well as "car computers" in catalogs, were a bit
pricey for me at the time. And I'm still not sure that *any* of them used an
actual fuel flow sensor, rather than a simple vacuum sensor or ignition timing
hookup.
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gull
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response 14 of 28:
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Jul 13 03:41 UTC 2000 |
Flat roads are scarce in this part of Minnesota, but if I find one I may try
it. I just have an idle curiousity as to how the figures would compare to
the Grand Am's.
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drew
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response 15 of 28:
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Jul 15 03:18 UTC 2000 |
I have uploaded ~drew/cardrag2.wk1, which now contains least-square code. Once
you have substituted your own data, and stretched or shrunk the column to fit
the number of points, you will need to execute a macro to get Lotus 123 to
do the matrix operations. The macro is activated by typing alt-M, and will
recalc the whole spreadsheet afterward.
A derivation of the least squares method can be found in ~drew/leastsq.doc
(MS Word format). I am confident that it is correct as the numbers it spits
out seem to agree with MATLAB's results.
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n8nxf
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response 16 of 28:
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Jul 18 02:35 UTC 2000 |
You could compensate for flatness and wind by running the test in both
directions. (Is the most efficient shape that of a falling rain drop?)
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rcurl
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response 17 of 28:
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Jul 18 06:39 UTC 2000 |
A falling rain drop looks like a pancake, somewhat thickened around the
periphery. It looks *nothing* like the metaphorical "teardrop". In fact,
it takes the shape that has nearly the maximum possible drag!
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n8nxf
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response 18 of 28:
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Jul 18 11:58 UTC 2000 |
Interesting! I didn't know that. So what shape has the least amount of
drag?
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rcurl
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response 19 of 28:
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Jul 18 16:07 UTC 2000 |
Something like the *proverbial* raindrop!
In regard to the shape of a real raindrop, consider that the maximum
pressure due to its fall velocity occurs at its 'nose'. Since the drop
is deformable, the nose gets pushed in, and the result is the deformed
pancake. This shape and air turbulence leads to the shape continually
and rapidly deforming. The drops you see fall so fast you can't see
all the bizarre shapes they take. Surface tension is fighting these
shape changes, so very small drops approach spheres and the shapes get
more irregular as the size increases, until the irregular motions
of the drops lead them to break up.
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gull
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response 20 of 28:
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Jul 18 18:01 UTC 2000 |
Most people assume the most aerodynamic shape looks like a rocket, with a
very sharp, tapered nose. The Concorde is another example of this shape.
However, rockets and the Concorde are both made to fly at *supersonic*
velocities. The most efficient shape at subsonic speeds has a rounded nose
and a tapered tail -- as rcurl says, much like a teardrop. Look at a 747;
blunt, rounded nose at the front, long, tapered tailcone at the rear. The
GM EV-1 is also a pretty good example; it has the lowest drag coefficent of
any production car, IIRC. Somewhere around 0.19. (It has only 19% the drag
of a flat plate the same size as its frontal area.) For comparison, most
pickup trucks have drag coefficients around 0.75.
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russ
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response 21 of 28:
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Jul 19 00:30 UTC 2000 |
Re #10: Considering that a falling raindrop is somewhere between
spherical (for the really small ones) to hamburger-shaped (for
the larger ones)... I'd guess not.
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drew
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response 22 of 28:
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Jul 19 17:11 UTC 2000 |
On drag coefficients, it should be noted that it is necessary to specify just
what dimensions you are counting as the "area" for the purpose. #20 specified
the frontal area. However, airfoils generally have drag - and lift -
coefficients listed based on the area of the top of the wing, ie, the large
surface.
The drag coefficient - or rather, that part of it that is not induced by lift,
is largely dependent on the Renolds number, another dimensionless quantity
based partly on the size of the object. Like with drag coefficients, the
Renolds number must be given in context of which dimensions of the object are
used as its size.
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gull
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response 23 of 28:
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Jul 19 18:20 UTC 2000 |
Very true. I was under the impression (possibly wrong) that for automobiles
the quoted drag coefficient is usually the frontal area. You need to know
both to get any idea of how "aerodynamic" the object really is. For example,
a 1960's VW bus has a better drag coefficent than a Beetle of the same era
-- the bus has a rounded nose and flat sides, and the Beetle has a flat
windshield and wide fenders that stick out into the airstream. The Beetle's
frontal area is so much smaller, though, that the total drag ends up being a
lot less. On flat ground the Beetle will do 80 with roughly the same amount
of power it takes to get the bus to do 60.
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rcurl
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response 24 of 28:
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Jul 19 23:49 UTC 2000 |
That's Reynolds number.
The lift is at right angles to the drag. By definition, drag and lift
are separate, orthogonal, forces. One can't say lift contributes to
drag, therefore. Of course, the whole profile contributes to both
drag and lift. The "projected frontal area" convention for a drag
coefficient is just that - a convention to define the drag coeff.
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